In epsilon delta method, is this possible? When we use the epsilon delta method, we use the Strictly inequality as follows:
function $f$ is continous at $x=a$ if and only if
For arbitrary $\epsilon$, there exist a $\delta$ such that 
If $ | x-a | < \delta$ then $ | f(x) - f(a) | < \epsilon$
Is it possible to change the strictly inequality($<$ to inequality($\leq$) in the above expression?
If possible is this equivalent?
 A: Yes, it is possible, and is equivalent. For a function $f \colon \def\R{\mathbf R}\R \to \R$ and $a \in \R$, we will show that 
$$ \forall \epsilon > 0\, \exists \delta > 0\, \forall x \in \R\, |x-a| < \delta \implies |f(x) - f(a)| < \epsilon \tag+$$
and 
$$ \forall \epsilon > 0\, \exists \delta > 0\, \forall x \in \R\, |x-a| \le \delta \implies |f(x) - f(a)| \le \epsilon \tag{$*$}$$
Suppose $(+)$ holds, to prove $(*)$ let $\epsilon > 0$. Using $(+)$ we get a $\delta' > 0$ such that 
$$ |x-a| < \delta' \implies |f(x) - f(a)| < \epsilon $$
Let $\delta := \frac{\delta'}2$. If $x \in \R$ is given with $|x-a| \le \delta$, then $|x-a| < \delta'$, hence $|f(x)-f(a)| < \epsilon$, which implies $|f(x)-f(a)| \le \epsilon$.
Now suppose $(*)$ holds and let $\epsilon > 0$. Apply $(*)$ with $\frac \epsilon2$ to obtain $\delta > 0$ with 
$$ |x-a| \le \delta \implies |f(x) - f(a)| \le \frac\epsilon2 $$
If now $x \in \R$ with $|x-a| < \delta$ is given, then $|x-a| \le \delta$, hence, by the choice of $\delta$, $|f(x)-f(a)| \le \frac\epsilon2< \epsilon$.
A: Surely it is. Since the statement holds for all $\epsilon > 0$, then if we take fixed $\epsilon>0$, then clearly $\frac{\epsilon}{2}>0$ so the statement $|f(x)-f(a)|\leqslant\frac{\epsilon}{2}<\epsilon$ is true as well (but $\delta$ in definition will probably be even smaller than before).
The reverse implication $<\implies \leqslant$ is obvious.
